What is the proper way to turn off a positive voltage supply with a NPN transistor?

The simplest way to control the positive rail is with a PNP.

Emitter to positive rail, base to control signal via a resistor (a typical value is [V+ - 0.5] / [Iload / 50] but it depends on the transistor - check Hfe / DC current gain) and collector to positive side of load.

Note it is just as easy to use a P-channel FET for this, with perhaps better results due to lower voltage drop across the transistor (at least at 200mA). There are some really nice P-Channel devices around nowadays with Rds[on] of 0.03 ohm with Vgs = -4V.

Not knowing the value of your positive rail makes it difficult to specify anything particular.

Unless you are willing to sacrifice about 1V or so of your positive rail, you can't do this with an NPN - there would have to be extra circuitry at that.

Cheers

PeteS

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If you have a load which is tied to the positive supply you can use an NPN as a low-side switch like this: (View in Courier)

+V | [LOAD] | ___ C ON/OFF>---[R]---B NPN E | GND

If you have a load which is tied to ground, you can use a PNP as a high-side switch like this:

+V | | __ E ON/OFF>---[R]--B PNP C | [LOAD] | GND

or like this:

+V +V | | [R] | | E +--[R]--B PNP | C ___ C | ON/OFF---[R]---B NPN [LOAD] E | | | GND GND

-- John Fields Professional Circuit Designer

Or else:

Your +V supply for the load can be higher than your ON/OFF switching voltage supply, too.

R1 is mostly there to make sure that Q2 goes off and stays off when Q1 becomes inactive. There are some small capacitances to overcome and, if it weren't present, the node it connects to would be fairly high impedance and may then respond to things like touching your finger to it. It needs to be smaller in value if you plan to run the switch faster, but for most uses it can be something like 5k, 10k, 22k or perhaps even 47k. The value isn't usually usually a critical one and is often in the range I mentioned. Q1 has to pull R1 down, though, when Q1 goes active. So making R1 smaller means that Q1 has to supply more active current. Making R1 larger reduces this problem, but means that switching gets slower and the node becomes more sensitive to things like touching it. 10k is reasonable for R1.

R2 needs to be set so as to supply the necessary base current drive for Q2, when active, _and_ any current that R1 requires at that time. The base current drive needed by Q2 will usually be somewhere in the range of 1/10th to 1/50th of the LOAD current. But planning on more current is safer than planning on less, so the figure of 1/10th is a very often-used estimate. Also, since the base-emitter voltage of Q2 will be something between .65V and perhaps .9V (with .8V being a reasonable guess lacking better knowledge), this means that R1's active current will be about .8V/R1. With a value for R1 in mind, you can compute this. These two currents are added together, so that you have I(R2) = I(LOAD)/10 + 0.8V/R1. To compute R2, though, you also need to know the voltage across it. To get that, just estimate the control voltage you are using (say V(on) is 3.3V or 5V or whatever it is) and then subtract about 0.7V from it. That will be the voltage at the emitter of Q1, which is the voltage across R2. So you then have the equation, R2 = (V(on) - 0.7) / (I(LOAD)/10 + 0.8V/R1). Finding a standard resistor value nearby this value is the next step. It can be a little smaller (driving Q2 harder) or it can be a little larger (eating into some of the conservative nature of the 1/10th estimate of Q2's base drive beta) and things should work okay.

There is also another approach that uses the same number of parts, too. It looks like:

This inverts the sense of the control line and it will require more current sinking capability for your control line driver. Which brings up another point, you need to know what currents your control line driver is capable/comfortable with. And if your LOAD requires a lot of current, this last complementary driver circuit may require too much from your control line driver and need yet another BJT to help out.

Anyway, keep in mind that I'm not a designer and cannot give you a professional quality comparison between the circuits I'm suggesting and the one that John suggested at the bottom of his post (included above.) One obvious difference is that my suggestion uses one less resistor to get the job done. I can think of a few other differences to comment on, but I'd rather let someone well informed make those points.

Jon